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  Why are the Yang-Mills instantons classified by the 3rd homotopy group of $S^3$?

+ 7 like - 0 dislike

This is a question that I would like to understand to some depth. I read in my QFT3 notes that instantons in Yang-Mills theories are classified by $\pi_3(S^3)$ without much more explanation. I would like to understand the following now that I am more familiar with some algebraic topology:

  • Why $\pi_3(S^3)$? We tried to see if we can classify them by the fundamental group and the second homotopy group of $S^3$ and there was no success?
  • Then, do the homotopy classes of the third homotopy group of $S^3$ encode all information needed? This means, are all instantons classified by the topological invariant $n$ homeomorphic to each other to a maximal degree? How can we be sure that there exist no other non-trivialities within a homotopy class that is not described by homotopy?
  • Also, why $S^3$? Is it related to the dimensionality of the space-time somehow? If not, does the dimensionality of space-time relate somehow to the fundamental and higher homotopy groups? 
  • What happens with supersymmetric Yang-Mills instantons?
  • In gauge/gavity dualities, in cases where the dual theory is some SYM (and not conformal) there are instantons. Their classification corresponds to something in the gravity (string) side?  

Ok, I guess I might have asked a lot. I hope that not everything is trivial to everybody else!

asked Nov 23, 2014 in Theoretical Physics by conformal_gk (3,625 points) [ no revision ]

A good question and it led to some very good answers. I just want to make it clear for future readers right up at the top here that Yang-Mills instantons are not, in general, classified by \(\pi_3(S^3)\). That classification requires two assumptions: (1) that the dimensionality of spacetime is \(d=4\); (2) that the gauge group associated with the Yang-Mills field in question is \(SU(2)\). Without the second assumption we have the more general statement: the instantons in \(d=4\) are classified by \(\pi_3(G)\), i.e., the third homotopy group of the gauge group \(G\).  Please correct me if I've said anything grossly incorrect (I'm still a Philistine when it comes to much precise mathematical terminology).

2 Answers

+ 5 like - 0 dislike

There are a few steps to arrive at that conclusion. Let $G$ denote a Lie group.

  1. Change signature to an Euclidean one. So one is dealing with a field theory in four-dimensional Euclidean space.
  2. Let us look for solutions to the equations of motion that lead to a finite action. A necessary condition is  that the field strength vanish as we go towards infinity. Let $B^4_R$ denote the ball of radius $R$. We shall model $\mathbb{R}^4$ as being obtain the limit $R\rightarrow \infty$. The boundary of the ball is $S^3_R$. It is necessary that  $F=O(R^{-2-\epsilon})$ for some  $\epsilon>0$ for a finite  action. 
  3. This implies that at the three-sphere at infinity, $S^3_\infty$, the gauge field must be pure gauge i.e., $\mathcal{A}_{\mu} = g^{-1} \partial_\mu g$, for some $g\in G$. We thus obtain the map: $g: S^3_\infty \rightarrow G$ specifies the boundary conditions on the gauge field for finite action.
  4. The homotopy class of such maps (boundary conditions) is classified by $\pi_3(G)$.

In summary, the finite action condition on the Euclidean action for Yang-Mills for the Lie group G implies that there exists inequivalent (under homotopy) class of solutions to the Euler-Lagrange equations of motion. The squaring trick shows that the action is bounded from below by a factor times $\int_{{\mathbb R}^4} F \wedge F$ and the bound is achieved when $F=\pm\ ^{*}F$. (I can provide more details if you wish). It is also known that the path-integral for $\mathcal{N}=2$ SYM theory localises on these configurations at weak coupling (Witten, late 80's) and to the Seiberg-Witten equations at strong coupling (Witten, Seiberg-Witten, around 1995).

I would strongly recommend that you read through Coleman's Erice lectures on Instantons (available as a chapter in his book "Aspects of Symmetry") to  understand things in more detail.

Comment added: I am not entirely convinced that the physics argument that I have given above is totally wrong as mentioned in nLab. It  would be nice to provide a connection possibly using the fact that $S^4=B^4/S^3$, where $B^4$ is the four-dimensional ball.  (Urs?)

answered Nov 24, 2014 by suresh (1,545 points) [ revision history ]
edited Nov 24, 2014 by suresh
+1(temporally retracted after reading the nlab post). Actually why do we need to set the context in Euclidean space? I imagine the $S^3$ results from the 1-point compactification in the spatial $R^3$ part.
+1 from me too. This is a good answer, but because of the nature of my question not complete. Still thanks a lot! Now, I do know that instantons are solutions of the classical eoms at infinity and that we take them to be in pure gauge. Furthermore, I guess that $S^3$ is explained as $\mathbb{R}^3 \cup \{\infty\}$ since we want these solutions to be non-vanishing at infinity. Still, how can we be sure that the complete classification is given by $\pi_{3}(S^3)$? This is why I am asking for other possible non-trivialities. Or is there some proven theorem that answers to my question? The above do change for greater than 4 dimensions? Does at some point one need to consider of exotic spheres where a lot of things are completely different? Finally I would like to hear what do people think about such a classification in the context of gauge/gravity dualities!
Quick comment as I am in the middle of something else. Exotic spheres differ as differentiable manifolds but are identical topologically. So I suspect that there might be nothing new in that context.

To Jia's comment: Such maps where one compactifies $\mathbb{R}^3$ by adding a point at infinity appear in the context of the Skyrme model. Such models involve scalars and off-hand, I can't see if it leads to anything for gauge theories. It may have some significance in the Hamiltonian formulation for Yang-Mills. I need to think a bit more about it.
@suresh, ah, ok, your $S^3$ is the boundary of $R^4$, I misunderstood that.
What? $\mathbb{R}^4$ has no boundary.
@conformal_gk: that's the only way I can make sense of suresh's "three-sphere at infinity". But you are obviously right that $R^4$ has no boundary in normal mathematical sense, now I'm equally confused.

@conformal_gk: This nlab thread seems to address the issue?

+ 5 like - 0 dislike

Notice that the one-point compactification happens one step earlier, when one considers fields on \(\mathbb{R}^4\) whose field strength vanishes at infinity. Since the question is about purely topological aspects, this is equivalent to considering gauge fields on the one-point compactification of \(\mathbb{R}^4\)which is \(S^4\).

Now gauge fields mathematically are principal connections on principal/vector bundles. There is a connection on any bundle and a unique self-dual one, so we may concentrate on classifying the underlying principal bundles.

This is done by degree-1 Cech cohomology with coefficients in the sheaf of gauge-group valued functions. While in general this is hard, over spheres this is particularly simple: the clutching construction says that it is sufficient to consider the cover of the 4-sphere given by two hemispheres that overlap a little at the equator.

Since everything is topological, you may think of covering \(S^4\) by the big hemi-4-sphere which is just \(S^4-\{\mbox{north pole}\}\) (this is homeomorphic to \(\mathbb{R}^4\)) together with the small hemi-4-sphere which is a tiny 4-disk around the northpole. This indeed gives the picture which one has from the physics, where the northpole is the "point at infinity" and we are working with fields that vanish at infinity.

Now by Cech cohomology, any G-principal bundle on the 4-sphere is defined by its transition function on the double overlap of these two patches (the two hemi-4-spheres). That double overlap is homeomorphic to \(S^3 \times (0,\epsilon)\) hence to the 3-sphere times a little thickening coming from the fact that the two hemi-4-spheres don't just touch but are required to overlap a little bit.

But such transition functions \(g : S^3 \times (0,\epsilon) \to G\) are also defined only up to gauge transformation, and a standard argument shows that their equivalence class is given simply by their homotopy class \([g] : S^3 \to G\). This is equivalently an element in the third homotopy group \([g] \in \pi_3(G)\) and hence these are the elements that classify the G-instanton sectors.

Now for the special case that \(G = SU(2)\) it so happens that as spaces this is \(SU(2) \simeq S^3\)

And so that is the reason of why SU(2)-instantons on Minkowski space are classified by an instanton number in \(\pi_3(S^3)\simeq \mathbb{Z}\).

(By the way, the analogous argument with the 4-sphere replaced by the 2-sphere and SU(2) replaced by U(1) yields the magnetic charge quantization of the Dirac monopole.)

Finally, with the principal bundle constructed this way you want to equip it with a principal connection \(\nabla\), the actual gauge field. You may choose a self-dual connection if you care, but for the following any other connection will do, too. If you use the standard formula for adding a connection to a bundle given by transition functions as above, then it will automatically satisfy the condition that it becomes pure gauge as one approaches infinity. 

Given that, you may produce the familiar differential 4-form \(\langle F_\nabla \wedge F_\nabla\rangle\) which is locally given by the expression familiar from the physics textbooks \(\mathrm{tr}(F_A \wedge F_A)\). It is Chern-Weil theory which says that the integral of that 4-formover the 4-sphere equals the instanton number that we obtained from Cech cohomology above.

For more exposition of what is going on see these slides, where the YM-instanton appears on slide 13 (29 of 52).

answered Nov 24, 2014 by Urs Schreiber (6,095 points) [ revision history ]
edited Nov 24, 2014 by Urs Schreiber
This is a really good answer. Thanks Urs.

Thanks for the feedback. I should add that generally \(\pi_n(G) = \pi_{n+1}(B G)\), where \(BG \) is the classifying space for G-principal bundles,  and this is how a homotopy theorist would arrive at the above conclusion in one line: G-bundles on \(S^4\) are classified by \([S^4,B G] = \pi_4(B G) = \pi_3(G)\) .

I wonder if one-point compactification can be done for solitons that involves matter field, such as the ones in Abelian Higgs model (Abrikosov solutions), far away the scalar matter field has the behaviour of $\lim_{r\to \infty} \phi(x)=\text{constant}\times e^{in\theta}$. Is it still possible/needed to form the homotopy picture based on one-point compactification?

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